Question

A wire of length \(L\) and resistance \(R\) is stretched to twice its length; what is the new resistance?

• A. \(R\)
• B. \(2R\)
• C. \(3R\)
• D. \(4R\)

Hint:

When a wire is stretched without changing volume: \[ A \propto \frac{1}{L} \] Thus resistance varies as \(R \propto L^2\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a wire with initial length \(L_1\) and resistance \(R_1 = R\). This wire is stretched so its new length \(L_2\) is twice the original length. We need to find the new resistance \(R_2\), assuming the material and temperature do not change.
Step 2: Key Formula or Approach:
The resistance of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \] where \(\rho\) is the resistivity of the material, \(L\) is the length, and \(A\) is the cross-sectional area. A key principle is that when a wire is stretched, its volume remains constant.
Volume, \(V = L \times A = \text{constant}\)
Step 3: Detailed Explanation:
Let the initial parameters be \(L_1, A_1, R_1\) and the final parameters be \(L_2, A_2, R_2\).
We are given: \(L_2 = 2L_1\) and \(R_1 = R\).
Since the volume is constant, \(L_1 A_1 = L_2 A_2\). We can find the new area \(A_2\):
\[ L_1 A_1 = (2L_1) A_2 \implies A_2 = \frac{L_1 A_1}{2L_1} = \frac{A_1}{2} \] So, stretching the wire to twice its length halves its cross-sectional area.
Now, let's find the new resistance \(R_2\):
\[ R_2 = \rho \frac{L_2}{A_2} \] Substitute the expressions for \(L_2\) and \(A_2\):
\[ R_2 = \rho \frac{2L_1}{A_1/2} = \rho \frac{4L_1}{A_1} = 4 \left( \rho \frac{L_1}{A_1} \right) \] Since \(R_1 = \rho \frac{L_1}{A_1}\), we have:
\[ R_2 = 4R_1 = 4R \] Step 4: Final Answer:
The new resistance of the wire is \(4R\).